First slide

Projection Under uniform Acceleration

Question

A projectile is fired at an angle of 45° with the horizontal. Elevation angle of the projectile at its highest point as seen from the point of projection, is :

Easy

A

45°
B

60°
C

$\tan^{- 1} (\frac{1}{2})$
D

$\tan^{- 1} (\frac{\sqrt{3}}{2})$

Solution

$tanθ = \frac{H_{\max}}{R / 2} = \frac{\frac{u^{2} \sin^{2} 45 °}{2 g}}{\frac{2 u^{2} \sin 45 ° \cos 45 °}{2 g}}$
$\frac{\tan 45 °}{2} = \frac{1}{2}$
$θ = \tan^{- 1} (\frac{1}{2})$

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